l = [(x,x) for x in xrange(10000)] d = dict(l) from time import clock t0=clock() for i in d: t = i + d[i] t1=clock() for k,v in d.items(): t = k + v t2=clock() for k,v in d.iteritems(): t = k + v t3=clock() for k,v in zip(d.iterkeys(),d.itervalues()): t = k + v t4=clock() print t1-t0, t2-t1, t3-t2, t4-t3
将这段脚本运行5次，结果如下：
python test.py 0.00184039735833 0.003292977712 0.00214993552657 0.00311549755797 python test.py 0.00182356570728 0.003393425046 0.00234863111466 0.003215660817 python test.py 0.00185107108827 0.00324563495762 0.002111751563 0.00313479237748 python test.py 0.0018215130669 0.00320950848705 0.00215814608806 0.00322798225041 python test.py 0.002166356955 0.00391807994377 0.00207604047314 0.00322757172233
显然第一种方法效率最高，第三种方法略差一点但相差无几，方法二四性能就差得多
不过实际的差别不是太大，不必过于纠结
